A student reacts benzene,C6H6. with bromine. Br2. in an attempt to prepare bromo

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(A) First find the moles of both: Br2 = 159.81 g/mol --> 60g Br2 / 159.81 g/mol = 0.3754 mol Br2 C6H6 = 78.11 g/mol --> 60g C6H6 / 78.11 g/mol = 0.7682 mol C3H6 If the reaction is 1:1 then by default the Br2 is your limiting reagent, you should only be able to (theoretically) produce 0.3754 mol of bromobenzene. 0.3754 mol bromobenzene * 157.01 g / mol = 58.949 g bromobenzene (b) If actual yield was 56.7g then % yield = actual/theoretical % yield = 56.7 / 58.949 = 96.2%

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